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Three Intersecting Circles

Date: 07/14/99 at 00:37:46
From: Tom
Subject: 3 circles intersecting

There are two circles:

  O1   X^2+Y^2+4X-4Y-8 = 0 
  O2   X^2+Y^2-X-Y-2 = 0.  

O1 and O2 intersect at points P and Q. Another circle:

  O3   3X^2+3Y^2-13X+KY+L = 0,

passes through P, Q, and A (3,1).  What is L?

My solution:

1. I solved X in terms of Y by subtracting O2 from O1, and I got 
   5X-3Y-6 = 0 and X = (3Y+6)/5.

2. I plugged X into O2 to solve Y: [(3Y+6)]^2+Y^2-(3Y+6)/5-Y-2 = 0 and 
   Y = [1+5sqrt(15)]/17 or [1-5sqrt(15)]/17.

3. With the two values of Y, I substituted them to X, and solved for 
   X: X = (3Y+6)/5 = [21+3sqrt(15)]/17 or [21-3sqrt(15)]/17.

4. I took 1 set of X and Y and substituted them into O3: 

   3X^2+3Y^2-13X+KY+L=0 --> 

   +K[1+5sqrt(15)]/17 + L = 0.  

Simplifying, I got: 

   -105+15sqrt(15)+[1-5sqrt(15)]K+17L = 0

5. Since I have 2 unknowns K and L, I need another equation, so I 
   plugged A(3, 1) into O3 to get K = 9-L.

6. I substituted K into the equation in step 4 to get L = 6.

Even though I solved this question, I think my steps were too 
complicated and prone to calculation errors. Do you have better ways 
to solve this question?  

Thank you for your help.

Date: 07/14/99 at 14:47:35
From: Doctor Rob
Subject: Re: 3 circles intersecting

Thanks for writing to Ask Dr. Math.

Yes, there is a simpler way, although yours is very logical and
straightforward (and correct!).

The family of circles passing through the two points of intersection 
is given by

  A*(X^2+Y^2+4*X-4*Y-8) + (1-A)*(X^2+Y^2-X-Y-2) = 0.

Since the circle

   3*X^2 + 3*Y^2 - 13*X + K*Y + L = 0, or
   X^2 + Y^2 + (-13/3)*X + (K/3)*Y + (L/3) = 0,

is one of them, the expressions on the lefthand side must be 
identical. Thus the coefficients of every term must be equal, and you 
get a system of equations:

   A + (1-A) = 1,  (X^2 terms)
   A + (1-A) = 1,  (Y^2 terms)
   4*A - (1-A) = -13/3,  (X terms)
   -4*A - (1-A) = K/3,  (Y terms)
   -8*A - 2*(1-A) = L/3,  (constant terms).

Solve the third equation for A, and substitute the solution into the
last two equations to figure out K and L.

Notice that you didn't need the fact that (3,1) lies on this circle
(although it does).

It is not too hard to prove the first assertion I made above for any
two intersecting circles. Let

   O1:  (X-h)^2 + (Y-k)^2 - r^2 = 0,
   O2:  (X-H)^2 + (Y-K)^2 - R^2 = 0,
   O3:  (X-a)^2 + (Y-b)^2 - c^2 = 0,
   O4:  A*[(X-h)^2+(Y-k)^2-r^2] + (1-A)*[(X-H)^2+(Y-K)^2-R^2] = 0.

The equation O4 is that of a circle, for all A, and it is satisfied by
any pair (x,y) that satisfies the equations of O1 and O2. Thus O4
passes through both points of intersection, no matter the value of A.
Furthermore, any circle O3 passing through the intersections of O1 and
O2 has center (a,b) lying on the line through their two centers (h,k)
and (H,K). That means that (k-K)*(a-H) = (h-H)*(b-K). If you set

   (a-H)/(h-H) = A = (b-K)/(k-K)

(ignoring any fraction of the form 0/0), then O4 will have the 
equation O3 for that particular A, so O3 is one of that family. In 
fact, you can solve for a and b in terms of h, k, H, K, and A:

   a = A*h + (1-A)*H,
   b = A*k + (1-A)*K.

This gives the one-to-one correspondence between values of A in O4 and
centers of circles (a,b) in O3.

Another way to look at this is to realize that the center of O3 lies
on the line between the centers of O1 and O2, (-2,2) and (1/2,1/2),
and, from O3's equation, completing the square, also lies on the line
X = 13/6. That allows you to determine the center of O3 uniquely. Then 
K is -6 times the Y-coordinate of the center of O3. Its radius is the 
distance from the center to (3,1), so you can write down the equation 
of the circle and expand it to find L.

- Doctor Rob, The Math Forum   
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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